If $(1+x)^{n}=C_{0} + C_{1} x + C_{2} x^2+......C_{n} x^n$

Question

If $(1+x)^{n}=C_{0} + C_{1} x + C_{2} x^2+......C_{n} x^n$, prove that: $$\textrm {a}. C_{0}C_{n}+C_{1}C_{n-1}+..........+C_{n}C_{0}=\dfrac {{2n}!}{{n}! \cdot {n}!}$$ $$\textrm {b}. {C_{0}}^{2}+{C_{1}}^{2}+{C_{2}}^{2}+......+{C_{n}}^2=\dfrac {{2n}!}{{n}! \cdot {n}!}$$

What is the general approach to such problems?

Answer 1

$a.$

$$(1+x)^n(x+1)^n=(1+x)^{2n}$$

Compare the coefficients of $x^n$

$b.$

$$(1+x)^n(1+x)^n=(1+x)^{2n}$$

Compare the coefficients of $x^n$

Answer 2

I am not sure part a can be proved without using some facts about binomial coefficients (specifically that the coefficient of $x^{2n}$ in $(1+x)^{2n}$ is ${2n \choose n}$, but this is easy to show). So compare coefficients in $((1+x)^n)^2$ and $(1+x)^{2n}$.

Here is a nice way to show that the expression in b. is equal to the expression in a. More generally, suppose $(x+y)^n = C_0y^n + C_1y^{n-1}x + ... + C_iy^{n-1}x^i + ... + C_nx^n$. Now use $(x+y)^n = (y+x)^n$ to derive b. from a.

Answer 3

• Notice that $C_k = C_{n-k}$
• Also, $$(1+x)^n(1+x)^n = (1+x)^{2n}$$
• The coefficient of $x^n$ would of the right hand side would be $\binom{2n}{n}$.
• The corresponding coefficient on the left would be $$C_0C_n + C_1C_{n-1}+\ldots + C_n C_0$$